Let two cards are drawn at random from a pack of 52 playing cards. Let $\mathrm{X}$ be the number of aces…
Let two cards are drawn at random from a pack of 52 playing cards. Let $\mathrm{X}$ be the number of aces obtained. Then the values of $\mathrm{E}(\mathrm{X})$ is
$\frac{5}{13}$
$\frac{1}{13}$
$\frac{2}{13}$
$\frac{37}{221}$
Solution
' $\mathrm{X}$ ' can take values 0.1 .2 .
Probability of getting no ace card.
$=\frac{{ }^{48} \mathrm{C}_2}{{ }^{52} \mathrm{C}_2}=\frac{48 !}{2 ! 46 !} \times \frac{2 ! 50 !}{52 !}=\frac{188}{221}$
Probability of getting 1 ace card
$=\frac{{ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 \times 48}{52 !} 2 ! 50 !=\frac{32}{221}$
Probability of getting 2 ace cards
$\begin{aligned}
& =\frac{{ }^4 \mathrm{C}_2 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 !}{2 ! 2 !} \times \frac{2 ! 50 !}{52 !}=\frac{1}{221} \\
& \mathrm{E}(\mathrm{X})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\
& =(0)\left(\frac{188}{221}\right)+(1)\left(\frac{32}{221}\right)+(2)\left(\frac{1}{221}\right)=\frac{2}{13}
\end{aligned}$